Answer:
Assuming that you meant the final velocity of 50 m/s was reached in 10 s, the answer would be 5 m/s^2.
Explanation:
So we update that with the values that we have.
then simplify that using algebra to solve for a and we get 5 m/s^2
Answer:
a) t = 3.027 10⁻⁹ s
, b) y = 2.25 10⁻² m
Explanation:
We can solve this problem using the kinematic relations
a) as on the x-axis there is no relationship
vₓ = x / t
t = x / vₓ
We reduce the magnitudes to the SI system
x = 5.6 cm (1m / 100 vm) = 0.056 m
we calculate
t = 0.056 / 1.85 10⁷
t = 3.027 10⁻⁹ s
b) the time is the same for the two movements, on the y axis
y = v₀t + ½ a t²
as the beam leaves horizontal there is no initial vertical velocity
y = ½ a t²
let's calculate
y = ½ 5.45 10¹⁵ (3.027 10⁻⁹)²
y = 2.25 10⁻² m
The echo is weaker than the original sound since the echo has a smaller amplitude than the original sound because sound spreads and its intensity decreases with distance. For example, when the sound from a speaker is heard and is compared to its echo, the echo heard has a lower amplitude than the actual sound.
Answer:
False
Explanation:
When the location of the poles changes in the z-plane, the natural or resonant frequency (ω₀) changes which in turn changes the damped frequency (ωd) of the system.
As the poles of a 2nd-order discrete-time system moves away from the origin then natural frequency (ω₀) increases, which in turn increases damped oscillation frequency (ωd) of the system.
ωd = ω₀√(1 - ζ)
Where ζ is called damping ratio.
For small value of ζ
ωd ≈ ω₀
Answer:
depende de que fenómenos nos referimos de acuerdo al los cuerpos de formación puede aver movimiento contante
